Develop Reference

How to Pass JsonString As a Parameter in Android WebAPI

string sortbyObj ="{'SortName':'DueDate','SortType':'asc'}";
string FiltersObj = "{'StartTime':'1/8/2016','EndTime':'3/8/2018','Ovedue':false,'Status':-1,'Priority':-1,'AssignedTo':-9999,'FollowUpBy':-9999,'Mytasks':false,'StartsWith':'','Endswith':''}";
string apiUrl = MainActivity.GetData("http://" + MainActivity.IPAddress + "/JTASKTest/api/jtasks/GetItemList?uName=admin&tgId=1&sortbyObj='"+sortbyObj.ToString()+"'&FiltersObj='"+FiltersObj.ToString()+"'");
I am trying to send string as a Parameter But it is giving 'Input string Formate not correct' Exception . Please Help Me how can i send this two Strings as a parametrs Through WebApi

If your API recognizes the url encoded Json String then I would suggest you to use something like this
string url = "http://yourapi/chat?msg=" + HttpUtility.UrlEncode(msg);
Detail answer here
in your case:
string apiUrl = MainActivity.GetData("http://" + MainActivity.IPAddress +"/JTASKTest/api/jtasks/GetItemList?uName=admin&tgId=1&sortbyObj='"+ HttpUtility.UrlEncode(sortbyObj.ToString())+"'&FiltersObj='"+HttpUtility.UrlEncode(FiltersObj.ToString()));
but in this case you should have to make sure that your API understands the URL Encoded JSON

How would you format this URL, it is giving me "Illegal character in path at index 70"?

I'm getting "Illegal character in path at index 70". And the final URL on debugging is coming like:
http://dev.example.com/Service/MyService.svc/CheckEmail/{0}
But I want the URL to be like:
http://dev.example.com/Service/MyService.svc/CheckEmail/rashid
I'm little new in Android, how can I achieve my desired result? Any help with explanation will be appreciated.
Below is the code:
String baseUrl = "http://dev.example.com/Service/MyService.svc/";
String url = String.format("CheckEmail/{0}", name);
HttpGet httpGet = new HttpGet(baseUrl + url);

Java doesn't use {} syntax for String.format. You confused it with a C# language. Java uses printf-like %-syntax for arguments.
Se here for details. In your case you should use
String url = String.format("CheckEmail/%s", name);

Just try this way may help you
String url = String.format("CheckEmail/%s", name);

How to send a getForObject request with parameters Spring MVC

I have a method on the Server side which gives me information about an specific name registered in my database. I'm accessing it from my Android application.
The request to Server is done normally. What I'm trying to do is to pass parameter to the server depending on the name I want to get.
Here's my Server side method:
#RequestMapping("/android/played")
public ModelAndView getName(String name) {
System.out.println("Requested name: " + name);
........
}
Here's the Android request to it:
private Name getName() {
RestTemplate restTemplate = new RestTemplate();
// Add the String message converter
restTemplate.getMessageConverters().add(
new MappingJacksonHttpMessageConverter());
restTemplate.setRequestFactory(
new HttpComponentsClientHttpRequestFactory());
String url = BASE_URL + "/android/played.json";
String nome = "Testing";
Map<String, String> params = new HashMap<String, String>();
params.put("name", nome);
return restTemplate.getForObject(url, Name.class, params);
}
In the server side, I'm only getting:
Requested name: null
Is it possible to send parameters to my Server like this?

The rest template is expecting a variable "{name}" to be in there for it to replace.
What I think you're looking to do is build a URL with query parameters you have one of two options:
Use a UriComponentsBuilder and add the parameters by that
String url = BASE_URL + "/android/played.json?name={name}"
Option 1 is much more flexible though.
Option 2 is more direct if you just need to get this done.
Example as requested
// Assuming BASE_URL is just a host url like http://www.somehost.com/
URI targetUrl= UriComponentsBuilder.fromUriString(BASE_URL) // Build the base link
.path("/android/played.json") // Add path
.queryParam("name", nome) // Add one or more query params
.build() // Build the URL
.encode() // Encode any URI items that need to be encoded
.toUri(); // Convert to URI
return restTemplate.getForObject(targetUrl, Name.class);

Change
String url = BASE_URL + "/android/played.json";
to
String url = BASE_URL + "/android/played.json?name={name}";
because the map contains variables for the url only!

how to url encode in android?

I am using grid view for displaying image using xml parsing,i got some exception like
java.lang.IllegalArgumentException: Illegal character in path at
index 80:
http://www.theblacksheeponline.com/party_img/thumbspps/912big_361999096_Flicking
Off Douchebag.jpg
How to solve this problem? I want to display all kind of url,anybody knows please give sample code for me.
Thanks All

URL encoding is done in the same way on android as in Java SE;
try {
String url = "http://www.example.com/?id=123&art=abc";
String encodedurl = URLEncoder.encode(url,"UTF-8");
Log.d("TEST", encodedurl);
} catch (UnsupportedEncodingException e) {
e.printStackTrace();
}

Also you can use this
private static final String ALLOWED_URI_CHARS = "##&=*+-_.,:!?()/~'%";
String urlEncoded = Uri.encode(path, ALLOWED_URI_CHARS);
it's the most simple method

As Ben says in his comment, you should not use URLEncoder.encode to full URLs because you will change the semantics of the URL per the following example from the W3C:
The URIs
http://www.w3.org/albert/bertram/marie-claude
and
http://www.w3.org/albert/bertram%2Fmarie-claude
are NOT identical, as in the second
case the encoded slash does not have
hierarchical significance.
Instead, you should encode component parts of a URL independently per the following from RFC 3986 Section 2.4
Under normal circumstances, the only
time when octets within a URI are
percent-encoded is during the process
of producing the URI from its
component parts. This is when an
implementation determines which of the
reserved characters are to be used as
subcomponent delimiters and which can
be safely used as data. Once
produced, a URI is always in its
percent-encoded form.
So, in short, for your case you should encode/escape your filename and then assemble the URL.

You don't encode the entire URL, only parts of it that come from "unreliable sources" like.
String query = URLEncoder.encode("Hare Krishna ", "utf-8");
String url = "http://stackoverflow.com/search?q=" + query;

URLEncoder should be used only to encode queries, use java.net.URI class instead:
URI uri = new URI(
"http",
"www.theblacksheeponline.com",
"/party_img/thumbspps/912big_361999096_Flicking Off Douchebag.jpg",
null);
String request = uri.toASCIIString();

you can use below method
public String parseURL(String url, Map<String, String> params)
{
Builder builder = Uri.parse(url).buildUpon();
for (String key : params.keySet())
{
builder.appendQueryParameter(key, params.get(key));
}
return builder.build().toString();
}

I tried with URLEncoder that added (+) sign in replace of (" "), but it was not working and getting 404 url not found error.
Then i googled for get better answer and found this and its working awesome.
String urlStr = "http://www.example.com/test/file name.mp4";
URL url = new URL(urlStr);
URI uri = new URI(url.getProtocol(), url.getUserInfo(), url.getHost(), url.getPort(), url.getPath(), url.getQuery(), url.getRef());
url = uri.toURL();
This way of encoding url its very useful because using of URL we can separate url into different part. So, there is no need to perform any string operation.
Then second URI class, this approach takes advantage of the URI class feature of properly escaping components when you construct a URI via components rather than from a single string.

I recently wrote a quick URI encoder for this purpose. It even handles unicode characters.
http://www.dmurph.com/2011/01/java-uri-encoder/

URL encoding in Android

How do you encode a URL in Android?
I thought it was like this:
final String encodedURL = URLEncoder.encode(urlAsString, "UTF-8");
URL url = new URL(encodedURL);
If I do the above, the http:// in urlAsString is replaced by http%3A%2F%2F in encodedURL and then I get a java.net.MalformedURLException when I use the URL.

You don't encode the entire URL, only parts of it that come from "unreliable sources".
Java:
String query = URLEncoder.encode("apples oranges", Charsets.UTF_8.name());
String url = "http://stackoverflow.com/search?q=" + query;
Kotlin:
val query: String = URLEncoder.encode("apples oranges", Charsets.UTF_8.name())
val url = "http://stackoverflow.com/search?q=$query"
Alternatively, you can use Strings.urlEncode(String str) of DroidParts that doesn't throw checked exceptions.
Or use something like
String uri = Uri.parse("http://...")
.buildUpon()
.appendQueryParameter("key", "val")
.build().toString();

I'm going to add one suggestion here. You can do this which avoids having to get any external libraries.
Give this a try:
String urlStr = "http://abc.dev.domain.com/0007AC/ads/800x480 15sec h.264.mp4";
URL url = new URL(urlStr);
URI uri = new URI(url.getProtocol(), url.getUserInfo(), url.getHost(), url.getPort(), url.getPath(), url.getQuery(), url.getRef());
url = uri.toURL();
You can see that in this particular URL, I need to have those spaces encoded so that I can use it for a request.
This takes advantage of a couple features available to you in Android classes. First, the URL class can break a url into its proper components so there is no need for you to do any string search/replace work. Secondly, this approach takes advantage of the URI class feature of properly escaping components when you construct a URI via components rather than from a single string.
The beauty of this approach is that you can take any valid url string and have it work without needing any special knowledge of it yourself.

For android, I would use
String android.net.Uri.encode(String s)
Encodes characters in the given string as '%'-escaped octets using the UTF-8 scheme. Leaves letters ("A-Z", "a-z"), numbers ("0-9"), and unreserved characters ("_-!.~'()*") intact. Encodes all other characters.
Ex/
String urlEncoded = "http://stackoverflow.com/search?q=" + Uri.encode(query);

Also you can use this
private static final String ALLOWED_URI_CHARS = "##&=*+-_.,:!?()/~'%";
String urlEncoded = Uri.encode(path, ALLOWED_URI_CHARS);
it's the most simple method

try {
query = URLEncoder.encode(query, "utf-8");
} catch (UnsupportedEncodingException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}

you can use below methods
public static String parseUrl(String surl) throws Exception
{
URL u = new URL(surl);
return new URI(u.getProtocol(), u.getAuthority(), u.getPath(), u.getQuery(), u.getRef()).toString();
}
or
public String parseURL(String url, Map<String, String> params)
{
Builder builder = Uri.parse(url).buildUpon();
for (String key : params.keySet())
{
builder.appendQueryParameter(key, params.get(key));
}
return builder.build().toString();
}
the second one is better than first.

Find Arabic chars and replace them with its UTF-8 encoding.
some thing like this:
for (int i = 0; i < urlAsString.length(); i++) {
if (urlAsString.charAt(i) > 255) {
urlAsString = urlAsString.substring(0, i) + URLEncoder.encode(urlAsString.charAt(i)+"", "UTF-8") + urlAsString.substring(i+1);
}
}
encodedURL = urlAsString;